Estimation quiz!

[Werner]

Or, even simpler:

Since e^x = 1 + x + x^2/2 +…, e^x > 1 + x if x>0
Take x = pi/e - 1, x>0 because pi>e
e^(pi/e-1) > 1 + pi/e - 1
e^(pi/e) / e > pi/e
e^(pi/e) > pi
(e^(pi/e))^(e) > pi^(e)
e^pi > pi^e

Cheers, Werner